Solve 10/y-60/1+6y=20

$$\frac{ 10 }{ y } - \frac{ 60 }{ 1+6y } =20$$

Solve for y

$$y=\frac{\sqrt{13}-1}{12}\approx 0.217129273$$

$$y=\frac{-\sqrt{13}-1}{12}\approx -0.38379594$$

Steps Using the Quadratic Formula

Variable y cannot be equal to any of the values -\frac{1}{6},0 since division by zero is not defined. Multiply both sides of the equation by y\left(6y+1\right), the least common multiple of y,1+6y.

$$\left(6y+1\right)\times 10-y\times 60=20y\left(6y+1\right)$$

Use the distributive property to multiply 6y+1 by 10.

$$60y+10-y\times 60=20y\left(6y+1\right)$$

Use the distributive property to multiply 20y by 6y+1.

$$60y+10-y\times 60=120y^{2}+20y$$

Subtract 120y^{2} from both sides.

$$60y+10-y\times 60-120y^{2}=20y$$

Subtract 20y from both sides.

$$60y+10-y\times 60-120y^{2}-20y=0$$

Combine 60y and -20y to get 40y.

$$40y+10-y\times 60-120y^{2}=0$$

Multiply -1 and 60 to get -60.

$$40y+10-60y-120y^{2}=0$$

Combine 40y and -60y to get -20y.

$$-20y+10-120y^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-120y^{2}-20y+10=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -120 for a, -20 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\left(-120\right)\times 10}}{2\left(-120\right)}$$

Square -20.

$$y=\frac{-\left(-20\right)±\sqrt{400-4\left(-120\right)\times 10}}{2\left(-120\right)}$$

Multiply -4 times -120.

$$y=\frac{-\left(-20\right)±\sqrt{400+480\times 10}}{2\left(-120\right)}$$

Multiply 480 times 10.

$$y=\frac{-\left(-20\right)±\sqrt{400+4800}}{2\left(-120\right)}$$

Add 400 to 4800.

$$y=\frac{-\left(-20\right)±\sqrt{5200}}{2\left(-120\right)}$$

Take the square root of 5200.

$$y=\frac{-\left(-20\right)±20\sqrt{13}}{2\left(-120\right)}$$

The opposite of -20 is 20.

$$y=\frac{20±20\sqrt{13}}{2\left(-120\right)}$$

Multiply 2 times -120.

$$y=\frac{20±20\sqrt{13}}{-240}$$

Now solve the equation y=\frac{20±20\sqrt{13}}{-240} when ± is plus. Add 20 to 20\sqrt{13}.

$$y=\frac{20\sqrt{13}+20}{-240}$$

Divide 20+20\sqrt{13} by -240.

$$y=\frac{-\sqrt{13}-1}{12}$$

Now solve the equation y=\frac{20±20\sqrt{13}}{-240} when ± is minus. Subtract 20\sqrt{13} from 20.

$$y=\frac{20-20\sqrt{13}}{-240}$$

Divide 20-20\sqrt{13} by -240.

$$y=\frac{\sqrt{13}-1}{12}$$

The equation is now solved.

$$y=\frac{-\sqrt{13}-1}{12} y=\frac{\sqrt{13}-1}{12}$$

Steps for Completing the Square

Variable y cannot be equal to any of the values -\frac{1}{6},0 since division by zero is not defined. Multiply both sides of the equation by y\left(6y+1\right), the least common multiple of y,1+6y.

$$\left(6y+1\right)\times 10-y\times 60=20y\left(6y+1\right)$$

Use the distributive property to multiply 6y+1 by 10.

$$60y+10-y\times 60=20y\left(6y+1\right)$$

Use the distributive property to multiply 20y by 6y+1.

$$60y+10-y\times 60=120y^{2}+20y$$

Subtract 120y^{2} from both sides.

$$60y+10-y\times 60-120y^{2}=20y$$

Subtract 20y from both sides.

$$60y+10-y\times 60-120y^{2}-20y=0$$

Combine 60y and -20y to get 40y.

$$40y+10-y\times 60-120y^{2}=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$40y-y\times 60-120y^{2}=-10$$

Multiply -1 and 60 to get -60.

$$40y-60y-120y^{2}=-10$$

Combine 40y and -60y to get -20y.

$$-20y-120y^{2}=-10$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-120y^{2}-20y=-10$$

Divide both sides by -120.

$$\frac{-120y^{2}-20y}{-120}=-\frac{10}{-120}$$

Dividing by -120 undoes the multiplication by -120.

$$y^{2}+\left(-\frac{20}{-120}\right)y=-\frac{10}{-120}$$

Reduce the fraction \frac{-20}{-120} to lowest terms by extracting and canceling out 20.

$$y^{2}+\frac{1}{6}y=-\frac{10}{-120}$$

Reduce the fraction \frac{-10}{-120} to lowest terms by extracting and canceling out 10.

$$y^{2}+\frac{1}{6}y=\frac{1}{12}$$

Divide \frac{1}{6}, the coefficient of the x term, by 2 to get \frac{1}{12}. Then add the square of \frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$y^{2}+\frac{1}{6}y+\left(\frac{1}{12}\right)^{2}=\frac{1}{12}+\left(\frac{1}{12}\right)^{2}$$

Square \frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$y^{2}+\frac{1}{6}y+\frac{1}{144}=\frac{1}{12}+\frac{1}{144}$$

Add \frac{1}{12} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y^{2}+\frac{1}{6}y+\frac{1}{144}=\frac{13}{144}$$

Factor y^{2}+\frac{1}{6}y+\frac{1}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{1}{12}\right)^{2}=\frac{13}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(y+\frac{1}{12}\right)^{2}}=\sqrt{\frac{13}{144}}$$

Simplify.

$$y+\frac{1}{12}=\frac{\sqrt{13}}{12} y+\frac{1}{12}=-\frac{\sqrt{13}}{12}$$

Subtract \frac{1}{12} from both sides of the equation.

$$y=\frac{\sqrt{13}-1}{12} y=\frac{-\sqrt{13}-1}{12}$$